Answer:
The velocity is given as -B sin(ωt) while the acceleration is given as -C cos(ωt).
Explanation:
As the complete question is not given here, a search for the question is made and the complete question is as given in the attached image.
From the given data the equation of displacement x as it varies with time is given as
[tex]x=A cos(\omega t)[/tex]
Here
- A is the distance of the point at which it is released from the equilibrium position
- ω is the characteristic of the system measured in rad/sec.
Now in order to find the velocity, it is given as
[tex]v_x=\frac{dx}{dt}\\v_x=\frac{d(Acos(\omega t))}{dt}\\v_x=A\frac{d(cos(\omega t))}{dt}\\v_x=A\frac{-sin(\omega t)d(\omega t)}{dt}\\v_x=-Asin(\omega t)\omega\\[/tex]
Suppose that
[tex]B=A\omega[/tex]
So the equation becomes
[tex]v_x=-Asin(\omega t)\omega\\v_x=-Bsin(\omega t)\\[/tex]
So the velocity is given as -B sin(ωt).
Now the acceleration is given as
[tex]a_x=\frac{d(v_x)}{dt}\\a_x=\frac{d(-Bsin(\omega t))}{dt}\\a_x=-B\frac{d(sin(\omega t))}{dt}\\a_x=-B\frac{cos(\omega t)d(\omega t)}{dt}\\a_x=-Bcos(\omega t)\omega\\[/tex]
Suppose that
[tex]C=B\omega[/tex]
So the equation becomes
[tex]a_x=-Bcos(\omega t)\omega\\a_x=-Ccos(\omega t)\\[/tex]
the acceleration is given as -C cos(ωt).
